Đáp án:
6) x=-66
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne \pm 1\\
\dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2} - 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to {x^2} + 2x + 1 - {x^2} + 2x - 1 - 4 = 0\\
\to 4x - 4 = 0\\
\to x = 1\\
2)DK:x \ne \left\{ {1;2} \right\}\\
\dfrac{{4\left( {2 - x} \right) + 3\left( {x - 1} \right) + \left( {x - 1} \right)\left( {2 - x} \right)}}{{\left( {x - 1} \right)\left( {2 - x} \right)}} = 0\\
\to 8 - 4x + 3x - 3 - {x^2} + 3x - 2 = 0\\
\to - {x^2} + 2x + 3 = 0\\
\to \left( {3 - x} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
3)DK:x \ne \left\{ { - 4; - 1;0} \right\}\\
\dfrac{{x\left( {x + 4} \right) - \left( {x + 1} \right)\left( {x + 4} \right) + x\left( {x + 1} \right)}}{{x\left( {x + 1} \right)\left( {x + 4} \right)}} = 0\\
\to {x^2} + 4x - {x^2} - 5x - 4 + {x^2} + x = 0\\
\to 0 = - 4\left( l \right)\\
\to x \in \emptyset \\
4)DK:x \ne - 2\\
\dfrac{{x + 1 - x + 1}}{{x + 2}} = 0\\
\to \dfrac{2}{{x + 2}} = 0\left( {vô lý} \right)\\
\to x \in \emptyset \\
5)DK:x \ne \left\{ { - 3;3} \right\}\\
\dfrac{{\left( {x + 2} \right)\left( {x + 3} \right) + \left( {x - 2} \right)\left( {x - 3} \right) - 2{x^2} - 12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\\
\to {x^2} + 5x + 6 + {x^2} - 5x + 6 - 2{x^2} - 12 = 0\\
\to 0x = 0\left( {ld} \right)
\end{array}\)
⇒ Phương trình có vô số nghiệm \(\forall x \ne \left\{ { - 3;3} \right\}\)
\(\begin{array}{l}
6)\dfrac{{x + 1}}{{65}} + 1 + \dfrac{{x + 3}}{{63}} + 1 = \dfrac{{x + 5}}{{61}} + 1 + \dfrac{{x + 7}}{{59}} + 1\\
\to \dfrac{{x + 66}}{{65}} + \dfrac{{x + 66}}{{63}} = \dfrac{{x + 66}}{{61}} + \dfrac{{x + 66}}{{59}}\\
\to \left( {x + 66} \right)\left( {\dfrac{1}{{65}} + \dfrac{1}{{63}} - \dfrac{1}{{61}} - \dfrac{1}{{59}}} \right) = 0\\
\to x + 66 = 0\\
\to x = - 66
\end{array}\)
